Lemma 115.5.1. Let $R$ be a ring and let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. If $t$ is integral over $R[x]$, then there exists an $\ell \geq 0$ such that for every $a \in R$ the element $\varphi (a)^\ell t$ is integral over $\varphi _ a : R[y] \to S$, defined by $y \mapsto \varphi (ax)$ and $r \mapsto \varphi (r)$ for $r\in R$.
115.5 Lemmas related to ZMT
The lemmas in this section were originally used in the proof of the (algebraic version of) Zariski's Main Theorem, Algebra, Theorem 10.123.12.
Proof. Say $t^ d + \sum _{i < d} \varphi (f_ i)t^ i = 0$ with $f_ i \in R[x]$. Let $\ell $ be the maximum degree in $x$ of all the $f_ i$. Multiply the equation by $\varphi (a)^\ell $ to get $\varphi (a)^\ell t^ d + \sum _{i < d} \varphi (a^\ell f_ i)t^ i = 0$. Note that each $\varphi (a^\ell f_ i)$ is in the image of $\varphi _ a$. The result follows from Algebra, Lemma 10.123.1. $\square$
Lemma 115.5.2. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Set $u_ n = \varphi (a_ n)$, $u_{n-1} = u_ n t + \varphi (a_{n-1})$, and so on till $u_1 = u_2 t + \varphi (a_1)$. Then all of $u_ n, u_{n-1}, \ldots , u_1$ and $u_ nt, u_{n-1}t, \ldots , u_1t$ are integral over $R$, and the ideals $(\varphi (a_0), \ldots , \varphi (a_ n))$ and $(u_ n, \ldots , u_1)$ of $S$ are equal.
Proof. We prove this by induction on $n$. As $u_ n = \varphi (a_ n)$ we conclude from Algebra, Lemma 10.123.1 that $u_ nt$ is integral over $R$. Of course $u_ n = \varphi (a_ n)$ is integral over $R$. Then $u_{n - 1} = u_ n t + \varphi (a_{n - 1})$ is integral over $R$ (see Algebra, Lemma 10.36.7) and we have
\[ \varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_{n - 1})t^{n - 1} + u_{n - 1}t^{n - 1} = 0. \]
Hence by the induction hypothesis applied to the map $S' \to S$ where $S'$ is the integral closure of $R$ in $S$ and the displayed equation we see that $u_{n-1}, \ldots , u_1$ and $u_{n-1}t, \ldots , u_1t$ are all in $S'$ too. The statement on the ideals is immediate from the shape of the elements and the fact that $u_1t + \varphi (a_0) = 0$. $\square$
Lemma 115.5.3. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Let $J \subset S$ be an ideal such that for at least one $i$ we have $\varphi (a_ i) \not\in J$. Then there exists a $u \in S$, $u \not\in J$ such that both $u$ and $ut$ are integral over $R$.
Proof. This is immediate from Lemma 115.5.2 since one of the elements $u_ i$ will not be in $J$. $\square$
The following two lemmas are a way of describing closed subschemes of $\mathbf{P}^1_ R$ cut out by one (nondegenerate) equation.
Lemma 115.5.4. Let $R$ be a ring. Let $F(X, Y) \in R[X, Y]$ be homogeneous of degree $d$. Assume that for every prime $\mathfrak p$ of $R$ at least one coefficient of $F$ is not in $\mathfrak p$. Let $S = R[X, Y]/(F)$ as a graded ring. Then for all $n \geq d$ the $R$-module $S_ n$ is finite locally free of rank $d$.
Proof. The $R$-module $S_ n$ has a presentation
\[ R[X, Y]_{n-d} \to R[X, Y]_ n \to S_ n \to 0. \]
Thus by Algebra, Lemma 10.79.4 it is enough to show that multiplication by $F$ induces an injective map $\kappa (\mathfrak p)[X, Y] \to \kappa (\mathfrak p)[X, Y]$ for all primes $\mathfrak p$. This is clear from the assumption that $F$ does not map to the zero polynomial mod $\mathfrak p$. The assertion on ranks is clear from this as well. $\square$
Lemma 115.5.5. Let $k$ be a field. Let $F, G \in k[X, Y]$ be homogeneous of degrees $d, e$. Assume $F, G$ relatively prime. Then multiplication by $G$ is injective on $S = k[X, Y]/(F)$.
Proof. This is one way to define “relatively prime”. If you have another definition, then you can show it is equivalent to this one. $\square$
Lemma 115.5.6. Let $R$ be a ring. Let $F(X, Y) \in R[X, Y]$ be homogeneous of degree $d$. Let $S = R[X, Y]/(F)$ as a graded ring. Let $\mathfrak p \subset R$ be a prime such that some coefficient of $F$ is not in $\mathfrak p$. There exists an $f \in R$ $f \not\in \mathfrak p$, an integer $e$, and a $G \in R[X, Y]_ e$ such that multiplication by $G$ induces isomorphisms $(S_ n)_ f \to (S_{n + e})_ f$ for all $n \geq d$.
Proof. During the course of the proof we may replace $R$ by $R_ f$ for $f\in R$, $f\not\in \mathfrak p$ (finitely often). As a first step we do such a replacement such that some coefficient of $F$ is invertible in $R$. In particular the modules $S_ n$ are now locally free of rank $d$ for $n \geq d$ by Lemma 115.5.4. Pick any $G \in R[X, Y]_ e$ such that the image of $G$ in $\kappa (\mathfrak p)[X, Y]$ is relatively prime to the image of $F(X, Y)$ (this is possible for some $e$). Apply Algebra, Lemma 10.79.4 to the map induced by multiplication by $G$ from $S_ d \to S_{d + e}$. By our choice of $G$ and Lemma 115.5.5 we see $S_ d \otimes \kappa (\mathfrak p) \to S_{d + e} \otimes \kappa (\mathfrak p)$ is bijective. Thus, after replacing $R$ by $R_ f$ for a suitable $f$ we may assume that $G : S_ d \to S_{d + e}$ is bijective. This in turn implies that the image of $G$ in $\kappa (\mathfrak p')[X, Y]$ is relatively prime to the image of $F$ for all primes $\mathfrak p'$ of $R$. And then by Algebra, Lemma 10.79.4 again we see that all the maps $G : S_ d \to S_{d + e}$, $n \geq d$ are isomorphisms. $\square$
Remark 115.5.7. Let $R$ be a ring. Suppose that we have $F \in R[X, Y]_ d$ and $G \in R[X, Y]_ e$ such that, setting $S = R[X, Y]/(F)$ we have (1) $S_ n$ is finite locally free of rank $d$ for all $n \geq d$, and (2) multiplication by $G$ defines isomorphisms $S_ n \to S_{n + e}$ for all $n \geq d$. In this case we may define a finite, locally free $R$-algebra $A$ as follows:
as an $R$-module $A = S_{ed}$, and
multiplication $A \times A \to A$ is given by the rule that $H_1 H_2 = H_3$ if and only if $G^ d H_3 = H_1 H_2$ in $S_{2ed}$.
This makes sense because multiplication by $G^ d$ induces a bijective map $S_{de} \to S_{2de}$. It is easy to see that this defines a ring structure. Note the confusing fact that the element $G^ d$ defines the unit element of the ring $A$.
Lemma 115.5.8. Let $R$ be a ring, let $f \in R$. Suppose we have $S$, $S'$ and the solid arrows forming the following commutative diagram of rings Assume that $R_ f \to S'$ is finite. Then we can find a finite ring map $R \to S''$ and dotted arrows as in the diagram such that $S' = (S'')_ f$.
\[ \xymatrix{ & S'' \ar@{-->}[rd] \ar@{-->}[dd] & \\ R \ar[rr] \ar@{-->}[ru] \ar[d] & & S \ar[d] \\ R_ f \ar[r] & S' \ar[r] & S_ f } \]
Proof. Namely, suppose that $S'$ is generated by $x_ i$ over $R_ f$, $i = 1, \ldots , w$. Let $P_ i(t) \in R_ f[t]$ be a monic polynomial such that $P_ i(x_ i) = 0$. Say $P_ i$ has degree $d_ i > 0$. Write $P_ i(t) = t^{d_ i} + \sum _{j < d_ i} (a_{ij}/f^ n) t^ j$ for some uniform $n$. Also write the image of $x_ i$ in $S_ f$ as $g_ i / f^ n$ for suitable $g_ i \in S$. Then we know that the element $\xi _ i = f^{nd_ i} g_ i^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} g_ i^ j$ of $S$ is killed by a power of $f$. Hence upon increasing $n$ to $n'$, which replaces $g_ i$ by $f^{n' - n}g_ i$ we may assume $\xi _ i = 0$. Then $S'$ is generated by the elements $f^ n x_ i$, each of which is a zero of the monic polynomial $Q_ i(t) = t^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} t^ j$ with coefficients in $R$. Also, by construction $Q_ i(f^ ng_ i) = 0$ in $S$. Thus we get a finite $R$-algebra $S'' = R[z_1, \ldots , z_ w]/(Q_1(z_1), \ldots , Q_ w(z_ w))$ which fits into a commutative diagram as above. The map $\alpha : S'' \to S$ maps $z_ i$ to $f^ ng_ i$ and the map $\beta : S'' \to S'$ maps $z_ i$ to $f^ nx_ i$. It may not yet be the case that $\beta $ induces an isomorphism $(S'')_ f \cong S'$. For the moment we only know that this map is surjective. The problem is that there could be elements $h/f^ n \in (S'')_ f$ which map to zero in $S'$ but are not zero. In this case $\beta (h)$ is an element of $S$ such that $f^ N \beta (h) = 0$ for some $N$. Thus $f^ N h$ is an element of the ideal $J = \{ h \in S'' \mid \alpha (h) = 0 \text{ and } \beta (h) = 0\} $ of $S''$. OK, and it is easy to see that $S''/J$ does the job. $\square$
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